Dirichlet and Liouville integral formulas

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Compute the Dirichlet integral

∭Vxpyqzr(1−x−y−z)s dx dy dz(p⩾0,q⩾0,r⩾0,s⩾0), \iiint \limits _{V} x^{p}y^{q}z^{r}(1-x-y-z)^{s}\, dx\, dy\, dz \quad (p\geqslant 0, q\geqslant 0, r\geqslant 0, s\geqslant 0) ,

where VV is the region bounded by the planes x+y+z=1x+y+z=1, x=0x=0, y=0y=0, z=0z=0, given the substitutions

x+y+z=ξ,y+z=ξη,z=ξηζ. x+y+z=\xi , \quad y+z=\xi \eta , \quad z=\xi \eta \zeta .
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Find

In=∬x1⩾0, x2⩾0, …, xn⩾0x1+x2+⋯+xn⩽a⋯∫dx1 dx2⋯dxn. I_{n} = \iint \limits _{\substack {x_{1}\geqslant 0,\ x_{2}\geqslant 0,\ \ldots ,\ x_{n}\geqslant 0 \\ x_{1}+x_{2}+\cdots +x_{n}\leqslant a}}\cdots \int dx_{1}\, dx_{2}\cdots dx_{n} .
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Find

∬x1⩾0, x2⩾0, …, xn⩾0x1+x2+⋯+xn⩽1⋯∫x1+x2+⋯+xn dx1 dx2⋯dxn. \iint \limits _{\substack {x_{1}\geqslant 0,\ x_{2}\geqslant 0,\ \ldots ,\ x_{n}\geqslant 0 \\ x_{1}+x_{2}+\cdots +x_{n}\leqslant 1}}\cdots \int \sqrt{x_{1}+x_{2}+\cdots +x_{n}}\, dx_{1}\, dx_{2}\cdots dx_{n} .
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Find the volume of the nn-dimensional simplex

x1a1+x2a2+⋯+xnan⩽1,xi⩾0(i=1,2,…,n)(ai>0, i=1,2,…,n). \dfrac {x_{1}}{a_{1}} + \dfrac {x_{2}}{a_{2}} + \cdots + \dfrac {x_{n}}{a_{n}} \leqslant 1 , \quad x_{i}\geqslant 0 \quad (i=1,2,\ldots ,n) \quad (a_{i}>0,\ i=1,2,\ldots ,n) .
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Prove the Dirichlet formula

∬x1,x2,…,xn⩾0x1+x2+⋯+xn⩽1⋯∫x1p1−1x2p2−1⋯xnpn−1 dx1 dx2⋯dxn=Γ(p1)Γ(p2)⋯Γ(pn)Γ(p1+p2+⋯+pn+1)(p1,p2,…,pn>0). \iint \limits _{\substack {x_{1},x_{2},\ldots ,x_{n}\geqslant 0 \\ x_{1}+x_{2}+\cdots +x_{n}\leqslant 1}}\cdots \int x_{1}^{p_{1}-1}x_{2}^{p_{2}-1}\cdots x_{n}^{p_{n}-1}\, dx_{1}\, dx_{2}\cdots dx_{n} = \dfrac {\Gamma (p_{1})\Gamma (p_{2})\cdots \Gamma (p_{n})}{\Gamma (p_{1}+p_{2}+\cdots +p_{n}+1)} \quad (p_{1},p_{2},\ldots ,p_{n}>0) .
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Prove the Liouville formula

∬x1,x2,…,xn⩾0x1+x2+⋯+xn⩽1⋯∫f(x1+x2+⋯+xn) x1p1−1x2p2−1⋯xnpn−1 dx1 dx2⋯dxn=Γ(p1)Γ(p2)⋯Γ(pn)Γ(p1+p2+⋯+pn)∫01f(u) up1+p2+⋯+pn−1 du(p1,p2,…,pn>0), \begin{align} \iint \limits _{\substack {x_{1},x_{2},\ldots ,x_{n}\geqslant 0 \\ x_{1}+x_{2}+\cdots +x_{n}\leqslant 1}}\cdots \int f(x_{1}+x_{2}+\cdots +x_{n})\, x_{1}^{p_{1}-1}x_{2}^{p_{2}-1}\cdots x_{n}^{p_{n}-1}\, dx_{1}\, dx_{2}\cdots dx_{n} \\ & = \dfrac {\Gamma (p_{1})\Gamma (p_{2})\cdots \Gamma (p_{n})}{\Gamma (p_{1}+p_{2}+\cdots +p_{n})}\int _{0}^{1} f(u)\, u^{p_{1}+p_{2}+\cdots +p_{n}-1}\, du \quad (p_{1},p_{2},\ldots ,p_{n}>0) , \end{align}

where f(u)f(u) is a continuous function.

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Find the multiple integral ∫⋯∫dx1 dx2…dxn\displaystyle \int \cdots \int dx_{1} \, dx_{2}\ldots dx_{n}, taken over the region where all xk>0x_{k}>0 and x1+x2+…+xn⩽1x_{1}+x_{2}+\ldots +x_{n}\leqslant 1.

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Find the multiple integral ∫⋯∫x1 dx1 dx2…dxn\displaystyle \int \cdots \int x_{1} \, dx_{1} \, dx_{2}\ldots dx_{n}, taken over the region where xk>0x_{k}>0 and x1+x2+…+xn⩽1x_{1}+x_{2}+\ldots +x_{n}\leqslant 1.

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Find the multiple integral

∫∣x1∣+∣x2∣+…+∣xn∣<a⋯∫dx1 dx2…dxn. \int \limits _{\left|x_{1}\right|+\left|x_{2}\right|+\ldots +\left|x_{n}\right|<a}\cdots \int dx_{1} \, dx_{2}\ldots dx_{n}.
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If

un=∫x1+x2+…+xn<a0<x1,0<x2,…,0<xn⋯∫x1α1−1x2α2−1…xnαn−1 dx1 dx2…dxn, u_{n} = \int \limits _{\substack {x_{1}+x_{2}+\ldots +x_{n}<a \\ 0<x_{1},0<x_{2},\ldots ,0<x_{n}}}\cdots \int x_{1}^{\alpha _{1}-1}x_{2}^{\alpha _{2}-1}\ldots x_{n}^{\alpha _{n}-1} \, dx_{1} \, dx_{2}\ldots dx_{n},

then it is easy to prove that un(a)=aα1+α2+…+αnvnu_{n}(a)=a^{\alpha_{1}+\alpha_{2}+\ldots +\alpha_{n}}v_{n}, where vn=un(1)v_{n}=u_{n}(1). Using this and the equality

vn=∫01x1α1−1 dx1∫x22+x32+…+xn2<1−x120<x2,…,0<xn⋯∫x2α2−1x3α3−1…xnαn−1 dx2 dx3…dxn, v_{n} = \int \limits _{0}^{1} x_{1}^{\alpha _{1}-1} \, dx_{1} \int \limits _{\substack {x_{2}^{2}+x_{3}^{2}+\ldots +x_{n}^{2}<1-x_{1}^{2} \\ 0<x_{2},\ldots ,0<x_{n}}}\cdots \int x_{2}^{\alpha _{2}-1}x_{3}^{\alpha _{3}-1}\ldots x_{n}^{\alpha _{n}-1} \, dx_{2} \, dx_{3}\ldots dx_{n},

prove that

un(a)=Γ(α1)Γ(α2)…Γ(αn)Γ(1+α1+α2+…+αn)aα1+α2+…+αn. u_{n}(a) = \frac{\Gamma (\alpha _{1})\Gamma (\alpha _{2})\ldots \Gamma (\alpha _{n})}{\Gamma (1+\alpha _{1}+\alpha _{2}+\ldots +\alpha _{n})}a^{\alpha _{1}+\alpha _{2}+\ldots +\alpha _{n}}.
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Reduce to a single integral the multiple integral

∬⋯∫x1α1−1x2α2−1…xnαn−1F(x1+x2+…+xn) dx1 dx2…dxn, \iint \cdots \int x_{1}^{\alpha _{1}-1}x_{2}^{\alpha _{2}-1}\ldots x_{n}^{\alpha _{n}-1}F(x_{1}+x_{2}+\ldots +x_{n}) \, dx_{1} \, dx_{2}\ldots dx_{n},

taken over the region defined by the inequalities

0<x1,0<x2,…,0<xn,x1+x2+…+xn<a. 0<x_{1}, \quad 0<x_{2}, \quad \ldots , \quad 0<x_{n}, \quad x_{1}+x_{2}+\ldots +x_{n}<a.
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Remark.
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One approach is a change of variables by the formulas x1+x2+…+xn=ux_{1}+x_{2}+\ldots +x_{n}=u, x1=x1x_{1}=x_{1}, x2=x2x_{2}=x_{2}, x3=x3,…,xn−1=xn−1x_{3}=x_{3},\ldots ,x_{n-1}=x_{n-1}.

Compute the integral ∬Gxmynzpdxdydz\iint_{G} x^{m} y^{n} z^{p} d x d y d z, where G={x2+y2+z2⩽1}G = \left\{ x^{2}+y^{2}+z^{2} \leqslant 1\right\}, and m,n,pm, n, p are nonnegative integers.

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Express in terms of values of the Γ\Gamma-function the Dirichlet integral

∭Gxpyqzr(1−x−y−z)sdxdydz \iiint _{G} x^{p} y^{q} z^{r}(1-x-y-z)^{s} d x d y d z

where G={x>0,y>0,z>0,x+y+z<1},p>0,q>0,r>0,s>0G = \left\{ x > 0, y > 0, z > 0, x+y+z < 1\right\} , p > 0, q > 0, r > 0, s > 0.

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Let Sn={∑k=1nxk⩽1,xk⩾0,k=1,…,n}S_{n} = \left\{ \sum_{k = 1}^{n} x_{k} \leqslant 1, x_{k} \geqslant 0, k = 1, \ldots , n\right\}. Derive the Dirichlet formula (x=(x1;…;xn))\left(x = \left(x_{1} ; \ldots ; x_{n}\right)\right)

∫Snx1p1−1x2p2−1…xnpn−1dx=Γ(p1)Γ(p2)…Γ(pn)Γ(p1+p2+…+pn+1),pk>0,k=1,…,n \int _{S_{n}} x_{1}^{p_{1}-1} x_{2}^{p_{2}-1} \ldots x_{n}^{p_{n}-1} d x = \frac{\Gamma \left(p_{1}\right) \Gamma \left(p_{2}\right) \ldots \Gamma \left(p_{n}\right)}{\Gamma \left(p_{1}+p_{2}+\ldots +p_{n}+1\right)}, \quad p_{k} > 0, \quad k = 1, \ldots , n
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Let SnS_{n} be the same pyramid as in problem 191191, and let f(t)f(t) be a function continuous for t⩾0t \geqslant 0. Derive the Liouville formula

∫Snf(x1+…+xn)x1p1−1xxp2−1…xnpn−1dx1dx2…dxn==Γ(p1)Γ(p2)…Γ(pn)Γ(p1+p2+…+pn)∫01f(t)tp1+p2+…+pn−1dt,pk>0,k=1,…,n. \begin{array}{rl} \int _{S_{n}} & f\left(x_{1}+\ldots +x_{n}\right) x_{1}^{p_{1}-1} x_{x}^{p_{2}-1} \ldots x_{n}^{p_{n}-1} d x_{1} d x_{2} \ldots d x_{n} = \\ \quad & = \frac{\Gamma \left(p_{1}\right) \Gamma \left(p_{2}\right) \ldots \Gamma \left(p_{n}\right)}{\Gamma \left(p_{1}+p_{2}+\ldots +p_{n}\right)} \int _{0}^{1} f(t) t^{p_{1}+p_{2}+\ldots +p_{n}-1} d t, \quad p_{k} > 0, \quad k = 1, \ldots , n. \end{array}
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Let a0,…,ana_{0}, \ldots , a_{n} be positive numbers, b0=a1b_{0}=a_{1}, bk=ak+1−akb_{k}=a_{k+1}-a_{k} for k=1,…,n(an+1=a0)k=1, \ldots , n\left(a_{n+1}=a_{0}\right),

Wn={(x1,…,xn)∈Rn∣∑1⩽k⩽nxk<1,x1,…,xn>0}. W_{n}=\left\{ \left(x_{1}, \ldots , x_{n}\right) \in \mathbb {R}^{n} \mid \sum _{1 \leqslant k \leqslant n} x_{k}<1, x_{1}, \ldots , x_{n}>0\right\} .
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(a)

Prove that for any nonnegative and continuous function ff on (0,+∞)(0,+\infty ) the equality

∫…∫f(b0+∑k=1nbkx1…xk)x1n−1x2n−2…xn−1dx1…dxn==∫…∫f(a0+∑k=1n(ak−a0)xk)dx1…dxn \begin{aligned} \int \ldots \int f\left(b_{0}+\sum _{k=1}^{n} b_{k} x_{1} \ldots x_{k}\right) x_{1}^{n-1} x_{2}^{n-2} \ldots x_{n-1} d x_{1} \ldots d x_{n} & = \\ & =\int \ldots \int f\left(a_{0}+\sum _{k=1}^{n}\left(a_{k}-a_{0}\right) x_{k}\right) d x_{1} \ldots d x_{n} \end{aligned}

holds.

(b)

Compute the integrals

∫…∫(0,1)nx1n−1x2n−2…xn−1dx1…dxn(b0+b1x1+b2x1x2+…+bnx1…xn)n+1∫…∫dx1…dxn(a0+(a1−a0)x1+(a2−a0)x2+…+(an−a0)xn)n+1. \begin{gathered} \int \ldots \int _{(0,1)^{n}} \frac{x_{1}^{n-1} x_{2}^{n-2} \ldots x_{n-1} d x_{1} \ldots d x_{n}}{\left(b_{0}+b_{1} x_{1}+b_{2} x_{1} x_{2}+\ldots +b_{n} x_{1} \ldots x_{n}\right)^{n+1}} \\ \int \ldots \int \frac{d x_{1} \ldots d x_{n}}{\left(a_{0}+\left(a_{1}-a_{0}\right) x_{1}+\left(a_{2}-a_{0}\right) x_{2}+\ldots +\left(a_{n}-a_{0}\right) x_{n}\right)^{n+1}} . \end{gathered}

(continued).

\idotsintx1α1−1x2α2−1⋯xp−1αp−1−1(1−x1−x2−⋯−xp−1)αp−1 dx1 dx2⋯dxp−1=Γ(α1)Γ(α2)⋯Γ(αp)Γ(α1+α2+⋯+αp). \idotsint x_{1}^{\alpha _{1}-1} x_{2}^{\alpha _{2}-1} \cdots x_{p-1}^{\alpha _{p-1}-1} (1 - x_{1} - x_{2} - \cdots - x_{p-1})^{\alpha _{p}-1} \, dx_{1} \, dx_{2} \cdots dx_{p-1} = \frac{\Gamma (\alpha _{1}) \Gamma (\alpha _{2}) \cdots \Gamma (\alpha _{p})}{\Gamma (\alpha _{1}+\alpha _{2}+\cdots +\alpha _{p})} .
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Define (u,v)=T(s,t)(u, v) = T(s, t) on the strip

0<s<∞,0<t<1 0 < s < \infty , \quad 0 < t < 1

by setting u=s−st,v=stu = s-s t, v = s t. Show that TT is a 1−11-1 mapping of the strip onto the positive quadrant QQ in R2\mathbb {R}^{2}. Show that JT(s,t)=sJ_{T}(s, t) = s.

For x>0,y>0x > 0, y > 0, integrate

ux−1e−uvy−1e−v u^{x-1} e^{-u} v^{y-1} e^{-v}

over QQ, use Theorem 10.9 to convert the integral to one over the strip, and derive formula (96) of Chap. 8 in this way. (For this application, Theorem 10.9 has to be extended so as to cover certain improper integrals. Provide this extension).

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Let r1,…,rkr_{1}, \ldots , r_{k} be nonnegative integers, and prove that

∫Qkx1r1⋯xkrkdx=r1!⋯rk!(k+r1+⋯+rk)! \int _{\mathbf{Q}^{k}} x_{1}^{r_{1}} \cdots x_{k}^{r_{k}} d x = \frac{r_{1}!\cdots r_{k}!}{\left(k+r_{1}+\cdots +r_{k}\right)!}

Hint: Use Exercise 12, Theorems 10.9 and 8.20. Note that the special case r1=⋯=rk=0r_{1} = \cdots = r_{k} = 0 shows that the volume of QkQ^{k} is 1/k!1 / k!.

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Let f:R→[0,+∞]f: \mathbb {R} \rightarrow \left[0,+\infty \right] be Lk/B∗\mathscr {L}^{k} / \mathcal{B}^{*}-measurable, and let c1,…,ckc_{1}, \ldots , c_{k} denote positive real numbers. Writing x=(x1,…,xk)\boldsymbol {x}=\left(x_{1}, \ldots , x_{k}\right), show that x↦f(x1+⋯+xk)\boldsymbol {x} \mapsto f\left(x_{1}+\cdots +x_{k}\right) on Rk\mathbb {R}^{k} is Lk/B∗\mathscr {L}^{k} / \mathcal{B}^{*}-measurable and

∫(0,+∞)kf(x1+⋯+xk)∏i=1kxici−1 dλˉk(x)=∏i=1kΓ(ci)Γ(∑i=1kci)∫0∞f(t)t∑i=1kci−1 dλˉ(t). \int _{(0,+\infty )^{k}} f\left(x_{1}+\cdots +x_{k}\right) \prod _{i=1}^{k} x_{i}^{c_{i}-1} \, \mathrm{d}\bar\lambda _{k}(\boldsymbol {x})=\frac{\prod _{i=1}^{k} \Gamma \left(c_{i}\right)}{\Gamma \left(\sum _{i=1}^{k} c_{i}\right)} \int _{0}^{\infty } f(t) t^{\sum _{i=1}^{k} c_{i}-1} \, \mathrm{d}\bar\lambda (t) .
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Remark.
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Use the appropriate form of Fubini's Theorem to obtain

left side=∫(0,+∞)k−1∏i=1k−1xici−1∫0∞f(x1+⋯+xk)xkck−1 dλˉ(xk) dλˉk−1(x1,…,xk−1)=∫(0,+∞)k−1∏i=1k−1xici−1∫x1+⋯+xk−1∞f(t)(t−x1−⋯−xk−1)ck−1 dλˉ(t) dλˉk−1(x1,…,xk−1)=∫0∞f(t)∫A(∏i=1k−1xici−1)(t−∑i=1k−1xi)ck−1 dλˉk−1(x1,…,xk−1) dλˉ(t)=∫0∞f(t)t∑i=1kci−1∫B(∏i=1k−1yici−1)(1−y1−⋯−yk−1)ck−1 dλˉk−1(y1,…,yk−1) dλˉ(t)=I1×I2, \begin{aligned} & \text{left side} \\ & =\int _{(0,+\infty )^{k-1}} \prod _{i=1}^{k-1} x_{i}^{c_{i}-1} \int _{0}^{\infty } f\left(x_{1}+\cdots +x_{k}\right) x_{k}^{c_{k}-1} \, \mathrm{d}\bar\lambda (x_{k}) \, \mathrm{d}\bar\lambda _{k-1}\left(x_{1},\ldots ,x_{k-1}\right) \\ & =\int _{(0,+\infty )^{k-1}} \prod _{i=1}^{k-1} x_{i}^{c_{i}-1} \int _{x_{1}+\cdots +x_{k-1}}^{\infty } f(t) \left(t-x_{1}-\cdots -x_{k-1}\right)^{c_{k}-1} \, \mathrm{d}\bar\lambda (t) \, \mathrm{d}\bar\lambda _{k-1}\left(x_{1},\ldots ,x_{k-1}\right) \\ & =\int _{0}^{\infty } f(t) \int _{A} \left(\prod _{i=1}^{k-1} x_{i}^{c_{i}-1}\right) \left(t-\sum _{i=1}^{k-1} x_{i}\right)^{c_{k}-1} \, \mathrm{d}\bar\lambda _{k-1}\left(x_{1},\ldots ,x_{k-1}\right) \, \mathrm{d}\bar\lambda (t) \\ & =\int _{0}^{\infty } f(t) t^{\sum _{i=1}^{k} c_{i}-1} \int _{B} \left(\prod _{i=1}^{k-1} y_{i}^{c_{i}-1}\right) \left(1-y_{1}-\cdots -y_{k-1}\right)^{c_{k}-1} \, \mathrm{d}\bar\lambda _{k-1}\left(y_{1},\ldots ,y_{k-1}\right) \, \mathrm{d}\bar\lambda (t) \\ & =I_{1} \times I_{2}, \end{aligned}

where I1=∫0∞f(t)t∑i=1kci−1 dλˉ(t)I_{1}=\int_{0}^{\infty } f(t) t^{\sum_{i=1}^{k} c_{i}-1} \, \mathrm{d}\bar\lambda (t) and I2I_{2} is the (constant, free of ff) integral over BB. The second equality follows by writing t=x1+⋯+xkt=x_{1}+\cdots +x_{k}. The set AA in the third equality is the set of (x1,…,xk−1)\left(x_{1},\ldots ,x_{k-1}\right) with x1,…,xk−1>0x_{1},\ldots ,x_{k-1}>0 and x1+⋯+xk−1<tx_{1}+\cdots +x_{k-1}<t. The fourth equality follows by writing xi=tyix_{i}=t y_{i} for i=1,…,k−1i=1,\ldots ,k-1; the set BB is the set of (y1,…,yk−1)\left(y_{1},\ldots ,y_{k-1}\right) with y1,…,yk−1>0y_{1},\ldots ,y_{k-1}>0 and y1+⋯+yk−1<1y_{1}+\cdots +y_{k-1}<1. If f(z)=e−zf(z)=e^{-z} the above work and Fubini's Theorem show that the left side is ∫(0,+∞)ke−(x1+⋯+xk)∏i=1kxici−1 dλˉk(x)=∏i=1kΓ(ci)\int_{(0,+\infty )^{k}} e^{-\left(x_{1}+\cdots +x_{k}\right)} \prod_{i=1}^{k} x_{i}^{c_{i}-1} \, \mathrm{d}\bar\lambda_{k}(\boldsymbol {x})=\prod_{i=1}^{k} \Gamma \left(c_{i}\right).

But the same derivation shows that in this case the left side is also ∫0∞e−tt∑i=1kci−1 dλˉ(t)×I2=Γ(∑i=1kci)×I2\int_{0}^{\infty } e^{-t} t^{\sum_{i=1}^{k} c_{i}-1} \, \mathrm{d}\bar\lambda (t) \times I_{2}=\Gamma \left(\sum_{i=1}^{k} c_{i}\right) \times I_{2}.

Therefore, I2=(∏i=1kΓ(ci))/Γ(∑i=1kci)I_{2}=\left(\prod_{i=1}^{k} \Gamma \left(c_{i}\right)\right)/\Gamma \left(\sum_{i=1}^{k} c_{i}\right). Going back to general nonnegative ff in the derivation above gives the result.

Let ff be as in problem 10.3.7. Let b1,…,bkb_{1}, \ldots , b_{k} denote positive real numbers, and write xi=yi1/bix_{i}=y_{i}^{1/b_{i}} for i=1,…,ki=1, \ldots , k, obtaining the identity

∫(0,+∞)kf(x1b1+⋯+xkbk)∏i=1kxici−1 dλˉk(x)=∏i=1kΓ(cibi)(∏i=1kbi)(Γ(∑i=1kcibi))∫0∞f(t)t∑i=1kcibi−1 dλˉ(t). \int _{(0,+\infty )^{k}} f\left(x_{1}^{b_{1}}+\cdots +x_{k}^{b_{k}}\right) \prod _{i=1}^{k} x_{i}^{c_{i}-1} \, \mathrm{d}\bar\lambda _{k}(\boldsymbol {x})=\frac{\prod _{i=1}^{k} \Gamma \left(\frac{c_{i}}{b_{i}}\right)}{\left(\prod _{i=1}^{k} b_{i}\right) \left(\Gamma \left(\sum _{i=1}^{k} \frac{c_{i}}{b_{i}}\right)\right)} \int _{0}^{\infty } f(t) t^{\sum _{i=1}^{k} \frac{c_{i}}{b_{i}}-1} \, \mathrm{d}\bar\lambda (t) .
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Show, as in paragraph (4.4.8), that the beta function satisfies B(a,b)=Γ(a)Γ(b)/Γ(a+b)B(a, b) = \Gamma (a) \Gamma (b) / \Gamma (a+b).

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